1. Shifted gradients begin at a time other than between periods 1& 2Must use multiple factors to find P in year 0 Arithmetic

2. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397

3. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 0 1 2 3 10 45 60 65 70 95 The cash flow diagram is as follows: Cash Flow Diagram i=12%

4. First find P 2 for the gradient ($5) and its base amount ($60) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 0 1 2 3 10 45 P 2 =? 60 65 70 95 01238 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

5. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 0 1 2 3 10 45 P 0 =? 60 65 70 95 01238 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

6. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 Next, find P A for the $60 amounts of years 1 & 2: P A = 60(P/A,12%,2) = $101.41 0 1 2 3 10 45 P A =? 60 65 70 95 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

7. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 Next, find P A for the $60 amounts of years 1 & 2: P A = 60(P/A,12%,2) = $101.41 Finally, add P 0 & P A to get P T in year 0: A P T = P 0 + P A = $396.70 Answer is (d) 0 1 2 3 10 45 P T = ? 60 65 70 95 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

8. Shifted gradients begin at a time other than between periods 1& 2 Geometric Equation yields P for all cash flow(i.e. base amt is included)For negative gradient, change signs in front of both g’s P=A{1-[(1+g)/(1+i)] n /(i-g)}